# Java Program for Given a sorted and rotated array, find if there is a pair with a given sum

Given an array that is sorted and then rotated around an unknown point. Find if the array has a pair with a given sum ‘x’. It may be assumed that all elements in the array are distinct.

**Examples :**

Input: arr[] = {11, 15, 6, 8, 9, 10}, x = 16 Output: true There is a pair (6, 10) with sum 16 Input: arr[] = {11, 15, 26, 38, 9, 10}, x = 35 Output: true There is a pair (26, 9) with sum 35 Input: arr[] = {11, 15, 26, 38, 9, 10}, x = 45 Output: false There is no pair with sum 45.

We have discussed a O(n) solution for a sorted array (See steps 2, 3 and 4 of Method 1). We can extend this solution for rotated array as well. The idea is to first find the largest element in array which is the pivot point also and the element just after largest is the smallest element. Once we have indexes largest and smallest elements, we use similar meet in middle algorithm (as discussed here in method 1) to find if there is a pair. The only thing new here is indexes are incremented and decremented in rotational manner using modular arithmetic.

Following is the implementation of above idea.

## Java

`// Java program to find a pair with a given ` `// sum in a sorted and rotated array` `class` `PairInSortedRotated` `{` ` ` `// This function returns true if arr[0..n-1] ` ` ` `// has a pair with sum equals to x.` ` ` `static` `boolean` `pairInSortedRotated(` `int` `arr[], ` ` ` `int` `n, ` `int` `x)` ` ` `{` ` ` `// Find the pivot element` ` ` `int` `i;` ` ` `for` `(i = ` `0` `; i < n - ` `1` `; i++)` ` ` `if` `(arr[i] > arr[i+` `1` `])` ` ` `break` `;` ` ` ` ` `int` `l = (i + ` `1` `) % n; ` `// l is now index of ` ` ` `// smallest element` ` ` ` ` `int` `r = i; ` `// r is now index of largest ` ` ` `//element` ` ` ` ` `// Keep moving either l or r till they meet` ` ` `while` `(l != r)` ` ` `{` ` ` `// If we find a pair with sum x, we` ` ` `// return true` ` ` `if` `(arr[l] + arr[r] == x)` ` ` `return` `true` `;` ` ` ` ` `// If current pair sum is less, move ` ` ` `// to the higher sum` ` ` `if` `(arr[l] + arr[r] < x)` ` ` `l = (l + ` `1` `) % n;` ` ` ` ` `else` `// Move to the lower sum side` ` ` `r = (n + r - ` `1` `) % n;` ` ` `}` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `/* Driver program to test above function */` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `arr[] = {` `11` `, ` `15` `, ` `6` `, ` `8` `, ` `9` `, ` `10` `};` ` ` `int` `sum = ` `16` `;` ` ` `int` `n = arr.length;` ` ` ` ` `if` `(pairInSortedRotated(arr, n, sum))` ` ` `System.out.print(` `"Array has two elements"` `+` ` ` `" with sum 16"` `);` ` ` `else` ` ` `System.out.print(` `"Array doesn't have two"` `+ ` ` ` `" elements with sum 16 "` `);` ` ` `}` `}` `/*This code is contributed by Prakriti Gupta*/` |

**Output : **

Array has two elements with sum 16

The time complexity of the above solution is O(n). The step to find the pivot can be optimized to O(Logn) using the Binary Search approach discussed here.

**How to count all pairs having sum x?**

The stepwise algo is:

- Find the pivot element of the sorted and the rotated array. The pivot element is the largest element in the array. The smallest element will be adjacent to it.
- Use two pointers (say left and right) with the left pointer pointing to the smallest element and the right pointer pointing to largest element.
- Find the sum of the elements pointed by both the pointers.
- If the sum is equal to x, then increment the count. If the sum is less than x, then to increase sum move the left pointer to next position by incrementing it in a rotational manner. If the sum is greater than x, then to decrease sum move the right pointer to next position by decrementing it in rotational manner.
- Repeat step 3 and 4 until the left pointer is not equal to the right pointer or until the left pointer is not equal to right pointer – 1.
- Print final count.

Below is implementation of above algorithm:

## Java

`// Java program to find ` `// number of pairs with ` `// a given sum in a sorted ` `// and rotated array.` `import` `java.io.*;` ` ` `class` `GFG` `{` ` ` `// This function returns` `// count of number of pairs` `// with sum equals to x.` `static` `int` `pairsInSortedRotated(` `int` `arr[], ` ` ` `int` `n, ` `int` `x)` `{` ` ` `// Find the pivot element. ` ` ` `// Pivot element is largest ` ` ` `// element of array.` ` ` `int` `i;` ` ` `for` `(i = ` `0` `; i < n - ` `1` `; i++)` ` ` `if` `(arr[i] > arr[i + ` `1` `])` ` ` `break` `;` ` ` ` ` `// l is index of` ` ` `// smallest element.` ` ` `int` `l = (i + ` `1` `) % n; ` ` ` ` ` `// r is index of ` ` ` `// largest element.` ` ` `int` `r = i;` ` ` ` ` `// Variable to store` ` ` `// count of number` ` ` `// of pairs.` ` ` `int` `cnt = ` `0` `;` ` ` ` ` `// Find sum of pair ` ` ` `// formed by arr[l] ` ` ` `// and arr[r] and ` ` ` `// update l, r and ` ` ` `// cnt accordingly.` ` ` `while` `(l != r)` ` ` `{` ` ` `// If we find a pair with ` ` ` `// sum x, then increment ` ` ` `// cnt, move l and r to ` ` ` `// next element.` ` ` `if` `(arr[l] + arr[r] == x)` ` ` `{` ` ` `cnt++;` ` ` ` ` `// This condition is required ` ` ` `// to be checked, otherwise ` ` ` `// l and r will cross each ` ` ` `// other and loop will never ` ` ` `// terminate.` ` ` `if` `(l == (r - ` `1` `+ n) % n)` ` ` `{` ` ` `return` `cnt;` ` ` `}` ` ` ` ` `l = (l + ` `1` `) % n;` ` ` `r = (r - ` `1` `+ n) % n;` ` ` `}` ` ` ` ` `// If current pair sum ` ` ` `// is less, move to ` ` ` `// the higher sum side.` ` ` `else` `if` `(arr[l] + arr[r] < x)` ` ` `l = (l + ` `1` `) % n;` ` ` ` ` `// If current pair sum ` ` ` `// is greater, move ` ` ` `// to the lower sum side.` ` ` `else` ` ` `r = (n + r - ` `1` `) % n;` ` ` `}` ` ` ` ` `return` `cnt;` `}` ` ` `// Driver Code` `public` `static` `void` `main (String[] args) ` `{` ` ` `int` `arr[] = {` `11` `, ` `15` `, ` `6` `, ` `7` `, ` `9` `, ` `10` `};` ` ` `int` `sum = ` `16` `;` ` ` `int` `n = arr.length;` ` ` ` ` `System.out.println(` ` ` `pairsInSortedRotated(arr, n, sum));` `}` `}` ` ` `// This code is contributed by ajit` |

**Output:**

2

**Time Complexity:** O(n) **Auxiliary Space:** O(1)

This method is suggested by Nikhil Jindal.

**Exercise:**

1) Extend the above solution to work for arrays with duplicates allowed.

This article is contributed by **Himanshu Gupta**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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